My issue is this,
say $f: \Omega \mapsto \mathbb{R}$ is a measurable function, for which $\int f d\mu$ exists. What does it then mean to say that $f$ is finite almost everywhere?
Some material I looked at, suggests that it is same as implying that the set $\{ x\in \Omega: |f(x)| = \infty\}$ is contained in a null set, this however is meaningless isn't it ? since $f$ is never going to take the value $\pm \infty$, so no such $x$ can be in $\Omega$.
Then I thought that we're working with the extended real line, $\mathbb{\bar R}$, but now, something like $f(x) = \infty$ is integrable, but never finite.
Sorry if this was too long, where am I wrong?
If $f$ takes values in $\mathbb R$ the $f$ is finite everywhere, hence also almost everywhere whether or not $\int f d\mu$ exists. Suppose $f$ takes values in $[-\infty, \infty]$. Then we say that $\int f d\mu $ exists if either $f^{+}$ or $f^{-}$ is integrable. This does not guarantee that $f$ is finite almost eveywhere. For example we can have $f(x)=\infty$ for all $x$.
If $\int |f| d\mu$ is finite then $f$ is finite almost everywhere: $\int |f| d\mu \geq \int_{\{f=\infty\}} |f| d\mu$ so we must have $\mu \{x: f(x)=\infty\} =0$. Similarly, $\mu \{x: f(x)=-\infty\} =0$ and $\mu \{x: f(x) \notin \mathbb R\}=0$.